The vertex of the parabola of an equation of the form \(\displaystyle y = ax^{2}+ bx + c\)  has \(\displaystyle x\)-coordinate \(\displaystyle -\frac{b}{2a}\). Here, we substitute \(\displaystyle a = 1, b = 12\), to obtain \(\displaystyle x\)-coordinate

\(\displaystyle -\frac{b}{2a} = -\frac{12}{2 (1)} = -6\).

To find the \(\displaystyle y \:\)-coordinate, substitute this for \(\displaystyle x\):

\(\displaystyle y = x^{2} + 12x + 20\)

\(\displaystyle y = (-6)^{2} + 12(-6) + 20 = 36 -72 + 20 = -16\)

The vertex is \(\displaystyle (-6, -16)\).

To find the \(\displaystyle x\)-intercepts of the parabola, substitute 0 for \(\displaystyle y \,\) in the equation:

\(\displaystyle x^{2} + 12x + 20 = 0\)

\(\displaystyle (x+ 2)(x+10) = 0\)

Either

\(\displaystyle x+2=0\), in which case \(\displaystyle x= -2\),

or

\(\displaystyle x+10 = 0\), in which case \(\displaystyle x = -10\).

The \(\displaystyle x\)-intercepts are \(\displaystyle (-2,0)\) and \(\displaystyle (-10,0)\).

The line includes \(\displaystyle (-6, -16)\) and either \(\displaystyle (-2,0)\) or \(\displaystyle (-10,0)\), so we find the slope in each case using the slope formula.

If the line includes \(\displaystyle (-6, -16)\) and \(\displaystyle (-2,0)\):

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{0 - (-16)}{-2 - (-6)} = \frac{16}{4} = 4\).

If the line includes \(\displaystyle (-6, -16)\) and \(\displaystyle (-10,0)\):

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{0 - (-16)}{-10 - (-6)} = \frac{16}{-4} = -4\)

We choose the first case, since the line has positive slope. The line through \(\displaystyle (-6, -16)\) and \(\displaystyle (-2,0)\) has as its equation, using the point-slope form with \(\displaystyle m = 4\) and point \(\displaystyle (-2,0)\):

\(\displaystyle y-y_{1} = m (x-x_{1})\)

\(\displaystyle y-0= 4\left [ (x-(-2) \right ]\)

\(\displaystyle y = 4 (x+2)\)

\(\displaystyle y = 4x+8\)

The \(\displaystyle x\)-intercept of the line coincides with that of the graph of the quadratic equation, which is a horizontal parabola; to find the \(\displaystyle x\)-intercept of the parabola, substitute 0 for \(\displaystyle y\,\) in the quadratic equation:

\(\displaystyle x = y ^{2} - y - 12\)

\(\displaystyle x =0 ^{2} - 0- 12\)

\(\displaystyle x = -12\)

The \(\displaystyle x\)-intercept of the parabola, and of the line, is \(\displaystyle (-12,0)\).

The \(\displaystyle y\,\)-intercept of the line coincides with one of those of the parabola; to find the \(\displaystyle y\,\)-intercepts of the parabola, substitute 0 for \(\displaystyle x\) in the equation:

\(\displaystyle x = y ^{2} - y - 12\)

\(\displaystyle y ^{2} - y - 12 = 0\)

\(\displaystyle (y+3) (y-4)= 0\)

Either

\(\displaystyle y +3= 0\), in which case \(\displaystyle y = -3\),

or

\(\displaystyle y -4= 0\), in which case \(\displaystyle y = 4\).

The \(\displaystyle y\,\)-intercepts of the parabola are \(\displaystyle (0,-3)\) and \(\displaystyle (0,4)\), so the \(\displaystyle y\,\)-intercept of the line is one of these. We will examine both possibilities.

If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of the line, then the slope of the line is \(\displaystyle -\frac{b}{a}\). If the intercepts of the line are \(\displaystyle (-12,0)\) and \(\displaystyle (0,-3)\), the slope of the line is \(\displaystyle -\frac{-3}{-12} = -\frac{1}{4}\); if the intercepts are \(\displaystyle (-12,0)\) and \(\displaystyle (0,4)\), the slope is \(\displaystyle -\frac{4}{-12} = \frac{1}{3}\). We choose the latter, since we are looking for a line with positive slope; since its \(\displaystyle y \,\)-intercept is \(\displaystyle (0,4)\), then we can substitute \(\displaystyle m = \frac{1}{3}, b = 4\) in the slope-intercept form of the line, \(\displaystyle y = mx+b\), to get the correct equation, \(\displaystyle y = \frac{1}{3}x+4\).

The point of intersection of the two lines has as its coordinates the values of \(\displaystyle x\) and \(\displaystyle y\,\) that make both of the given linear equations true. Therefore, we seek to find the solution of the system of equations:

\(\displaystyle 7x-4y = 64\)

\(\displaystyle 4x+7y = 32\)

We need only find \(\displaystyle x\), so multiply both sides of the two equations by 7 and 4, respectively. Then add:

\(\displaystyle 49x-28y = 448\)

\(\displaystyle \underline{16x+28y = 128}\)

\(\displaystyle 65x\)              \(\displaystyle =576\)

\(\displaystyle x = 576 \div 65 \approx 8.9\),

making 9 the correct response.

The \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts of the line are, respectively, \(\displaystyle (3,0)\) and \(\displaystyle (0,5)\).  If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) are the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of a line, the slope of the line is \(\displaystyle -\frac{b}{a}\). This makes the slope of the green line \(\displaystyle -\frac{5}{3}\). 

Any line perpendicular to this line must have as its slope the opposite of the reciprocal of this, or \(\displaystyle \frac{3}{5}\). Since the desired line must also have \(\displaystyle y\,\)-intercept \(\displaystyle (0,5)\), then the slope-intercept form of the line is

\(\displaystyle y = mx+b\)

\(\displaystyle y = \frac{3}{5}x+ 5\)

which can be rewritten in standard form:

\(\displaystyle 5y =5 \left ( \frac{3}{5}x+ 5 \right )\)

\(\displaystyle 5y =3x+ 25\)

\(\displaystyle 5y - 5y - 25 =3x+ 25 - 5y - 25\)

\(\displaystyle 3x-5y = -25\)

The vertices of the triangle are the point of intersection of the graphs of the lines of the two equations, and the \(\displaystyle x\)-intercepts of those lines.

The \(\displaystyle x\)-intercept of the line of the equation \(\displaystyle 2x+ 4y= 9\) can be found by setting \(\displaystyle y = 0\) and solving for \(\displaystyle x\):

\(\displaystyle 2x+ 4y= 9\)

\(\displaystyle 2x+ 4 (0)= 9\)

\(\displaystyle 2x+0 = 9\)

\(\displaystyle 2x=9\)

\(\displaystyle \frac{2x}{2}=\frac{9}{2}\)

\(\displaystyle x = 4 \frac{1}{2}\)

The \(\displaystyle x\)-intercept of the line of the equation \(\displaystyle 4x - 2y = 13\) can be found the same way:

\(\displaystyle 4x - 2y = 13\)

\(\displaystyle 4x - 2 (0) = 13\)

\(\displaystyle 4x - 0 = 13\)

\(\displaystyle 4x= 13\)

\(\displaystyle \frac {4x}{4}= \frac {13}{4}\)

\(\displaystyle x= 3 \frac {1}{4}\)

The \(\displaystyle x\)-intercepts are \(\displaystyle \left ( 4 \frac{1}{2}, 0 \right )\) and \(\displaystyle \left ( 3 \frac {1}{4}, 0 \right )\); these are two of the vertices of the triangle, and since the segment connecting them is horizontal, this will be taken as the base. The length of the base is the difference of the \(\displaystyle x\)-coordinates:

\(\displaystyle b = 4 \frac{1}{2} -3 \frac {1}{4} = 1\frac {1}{4}\)

The intersection of the lines of the equations \(\displaystyle 2x+ 4y= 9\) and \(\displaystyle 4x - 2y = 13\) can be found by solving the system of linear equations, as follows:

\(\displaystyle 4x - 2y = 13\)

\(\displaystyle 2 \cdot (4x - 2y )= 2 \cdot 13\)

\(\displaystyle 8x - 4y = 2 6\)

\(\displaystyle \underline{2x+ 4y= 9}\)

\(\displaystyle 10x\)         \(\displaystyle = 35\)

\(\displaystyle \frac{10x}{10} = \frac{35}{10}\)

\(\displaystyle x = \frac{7}{2} = 3\frac{1}{2}\)

\(\displaystyle 2 \cdot \frac{7}{2}+ 4y= 9\)

\(\displaystyle 7+ 4y= 9\)

\(\displaystyle 7+ 4y- 7 = 9 - 7\)

\(\displaystyle 4y = 2\)

\(\displaystyle \frac{4y}{4} = \frac{2}{4}\)

\(\displaystyle y = \frac{1}{2}\)

The point of intersection, and the third vertex of the triangle, is \(\displaystyle \left ( 3\frac{1}{2}, \frac{1}{2} \right )\).

Since we are taking the horiztonal segment to be the base, the height will be the vertical distance to this third point - namely, the \(\displaystyle y\)-coordinate \(\displaystyle \frac{1}{2}\). The area is half the product of the base and the height:

\(\displaystyle A = \frac{1}{2} bh\)

\(\displaystyle =\frac{1}{2} \cdot 1 \frac {1}{4} \cdot \frac{1}{2}\)

\(\displaystyle =\frac{1}{2} \cdot \frac {5}{4} \cdot \frac{1}{2}\)

\(\displaystyle = \frac {5}{16}\)

The intercepts of the line of the equation \(\displaystyle 4x+ 7y = 50\) can be found by substituting 0 for each variable, in turn:

\(\displaystyle x\)-intercept:

\(\displaystyle 4x+ 7y = 50\)

\(\displaystyle 4x+ 7 (0) = 50\)

\(\displaystyle 4x+ 0= 50\)

\(\displaystyle 4x = 50\)

\(\displaystyle \frac{4x }{4}= \frac{50}{4}\)

\(\displaystyle x = \frac{25}{2}\)

The \(\displaystyle x\)-intercept is the point \(\displaystyle \left ( \frac{25}{2}, 0 \right )\)

\(\displaystyle y\)-intercept:

\(\displaystyle 4x+ 7y = 50\)

\(\displaystyle 4 (0)+ 7y = 50\)

\(\displaystyle 0+ 7y = 50\)

\(\displaystyle 7y = 50\)

\(\displaystyle \frac{7y }{7}= \frac{50}{7}\)

\(\displaystyle y= \frac{50}{7}\)

The \(\displaystyle y\)-intercept is the point \(\displaystyle \left ( \frac{50}{7} , 0 \right )\).

This line and the axes together form a right triangle. The horizontal leg is the segment connecting the origin to \(\displaystyle \left ( \frac{25}{2}, 0 \right )\), and its length is \(\displaystyle \frac{25}{2}\). The vertical leg is the segment connecting the origin to \(\displaystyle \left ( \frac{50}{7} , 0 \right )\), and its length is \(\displaystyle \frac{50}{7}\). The area of a right triangle is half the product of these legs, which is 

\(\displaystyle A = \frac{1}{2} \cdot \frac{50}{7} \cdot \frac{25}{2} = \frac{1,250}{28} = \frac{625}{14}= 44 \frac{9}{14}\).

Of the five responses, 45 comes closest to the correct area.